A) \[{{y}^{2}}=a\text{ }(a-2x)\]
B) \[{{x}^{2}}=a\text{ }(a-2y)\]
C) \[{{x}^{2}}+{{y}^{2}}={{(y-a)}^{2}}\]
D) None of these
Correct Answer: A
Solution :
\[T\equiv hx+ky-{{a}^{2}}=0\] \[\Rightarrow a=\frac{ah+0-{{a}^{2}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\] \[P(3,\,4)\]. Þ \[{{k}^{2}}=a(a-2h)\] \ The locus is\[{{y}^{2}}=a(a-2x)\].You need to login to perform this action.
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